$\begingroup$Try writing out the scheme as the multiplication of matrices rather than with the arrows. E.g. instead of $[\vec{v}]_\lambda\to C \to [\vec{v}]_\beta$, write $[\vec{v}]_\beta=C[\vec{v}_\lambda]$, and go from there.$\endgroup$
– AtaraxiaMay 24 '13 at 23:01

1 Answer
1

Let's call the standard basis vectors $e_1,\ e_2,\ e_3$. Then we have $\beta=(e_1,e_2,e_3)$ and $\lambda=(e_3,e_2,e_1)$.

Now you have luckily $T^{-1}=T$, as $T$ simply exchanges the basis vectors $e_1$ and $e_3$.

But, for this specific exercise, we can find the answer more directly:

We need $G'=M(g;\lambda,\lambda)$. Because $G\cdot e_i$ gives the $i$th column of the matrix $G$, we have
$$g(e_1)=G\cdot e_1 = e_1+6e_2$$
This means that the third column of $G'$ will be $\pmatrix{0\\6\\1}$, as these are the $\lambda$-coordinates of the $g$-image of the third basis vector of $\lambda$.